Mg+2HCl → MgCl2+H2

nMgCl2=4,25÷95=0,045 mol

⇒ nMg=0,045 mol

⇒ m=24×0,045=1,08g

nH2=0,045 mol

⇒ VH2=0,045×22,4=1,008g

nHCl=0,045×2=0,09 mol

⇒ mHCl=36,5×0,09=3,285g

⇒C% HCl=3,285÷100=0,03285%

C% MgCl2=4,25÷(100+1,08-0,045×2)=4,21%

MgCl2+2NaOH → Mg(OH)2 + 2NaCl

nMg(OH)2=0,045 mol=nMgCl2

⇒m kết tủa=58×0,045=2,61g

Em tham khảo nha :

 \(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{MgC{l_2}}} = \dfrac{{4,25}}{{95}} = 0,045mol\\
{n_{Mg}} = {n_{MgC{l_2}}} = 0,045mol\\
{m_{Mg}} = 0,045 \times 24 = 1,08g\\
b)\\
{n_{{H_2}}} = {n_{MgC{l_2}}} = 0,045mol\\
{V_{{H_2}}} = 0,045 \times 22,4 = 1,008l\\
c)\\
{n_{HCl}} = 2{n_{MgC{l_2}}} = 0,09mol\\
{m_{HCl}} = 0,09 \times 36,5 = 3,285g\\
C{\% _{HCl}} = \dfrac{{3,285}}{{100}} \times 100\%  = 3,285\% \\
d)\\
{m_{ddspu}} = 1,08 + 100 - 0,09 = 100,99g\\
C{\% _{MgC{l_2}}} = \dfrac{{4,25}}{{100,99}} \times 100\%  = 4.21\% \\
e)\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
{n_{Mg{{(OH)}_{ 2}}}} = {n_{MgC{l_2}}} = 0,045mol\\
{m_{Mg{{(OH)}_2}}} = 0,045 \times 58 = 2,61g
\end{array}\)